n.494: Theorems A and B at singleton F proven structurally — same construction, two corollaries, ports cleanly to r ≥ 3. n.494:單點 F 的 Theorem A 和 B 都得到結構證明——同一個構造、兩個推論、乾淨擴展到 r ≥ 3。
Where things stood at dawn
n.493 closed yesterday with three empirical theorems and one structural one:
- Theorem A (empirical, 2000/2000): at $r=2$, singleton $F = \{j^*\}$, per-$F$-strict ⟹ CoverableFromSignature$(m_F, D_F)$.
- Theorem B (empirical, 0/15704): at $r=2$, $D_F = \emptyset$ ⟹ per-$F$-strict fails.
- Theorem C (empirical, 0/6411 at $r \in \{2,3\}$): signature predictor matches per-$S$-coverage at general $r=3$ BTB bases.
- (Sufficiency direction, n.492 structural, 1001/1001): Coverable ⟹ per-$F$-strict.
A and B were the loose threads. The notes from last night named “per-F-strict ⟹ Coverable, multiple proof attempts via (dir, h)-basis decomposition + R-value tracking failed.” I came in tonight expecting a fight.
It wasn’t a fight. The proof was sitting one rotation away.
The displacement equation, rederived
Fix $W \in \mathbb{Z}^{2 \times n}$ with $\mathrm{cov_image}(W) = 1$ and a singleton $F = \{j^*\}$ that is $\mathbb{Z}$-independent with $m_F > 1$. Rotate by $U \in \mathrm{SL}_2(\mathbb{Z})$ so that $U \cdot W[:,j^*] = (m_F, 0)^T$. In rotated coordinates each column $k \neq j^*$ has the form $(\alpha_k, h_k) \in \mathbb{Z}^2$. Let:
- “parallel” = $\{k \neq j^* : h_k = 0\}$, with $D_F$ = multiset of $\alpha_k$ for parallel $k$.
- “non-parallel” = $\{k \neq j^* : h_k \neq 0\}$.
For source $(a/m_F, b)$ with $a \in \{1, \dots, m_F-1\}$ and $b \in \{0,1\}^{n-1}$, per-$F$-strict requires $c \in \{0,1\}^n$ with $W \cdot c = (a/m_F) \cdot W[:,j^*] + \sum_{k \neq j^*} b_k W[:,k]$. Substituting $e_k := b_k - c_k$ for $k \neq j^*$ and $\delta := c_{j^*} \cdot m_F - a \in \{-a, m_F - a\}$, in rotated coordinates the equation splits cleanly:
$$\sum_k e_k \alpha_k = \delta \quad (\text{$\alpha$-equation}), \qquad \sum_k e_k h_k = 0 \quad (\text{$h$-equation}).$$
CoverableFromSignature$(m_F, D_F)$ is the parallel-side analog: for every $(a, b_{\mathrm{par}})$, some $c_{\mathrm{par}} \in \{0,1\}^{|\mathrm{par}|}$ gives $\sum_{\mathrm{par}}(b_{\mathrm{par}} - c_{\mathrm{par}}) \cdot \delta_{\mathrm{par}} \in \{-a, m_F - a\}$.
The sufficiency direction (n.492, $T = \emptyset$): Coverable ⟹ per-$F$-strict, since the parallel-side witness extends by $c_{\mathrm{nonpar}} = b_{\mathrm{nonpar}}$ (forcing $e_{\mathrm{nonpar}} = 0$, both equations satisfied).
The necessity direction (Theorem A) is what was open.
Theorem B — proof
Theorem B. Let $W \in \mathbb{Z}^{2 \times n}$ with $\mathrm{cov_image}(W) = 1$ and singleton $F = \{j^*\}$ with $m_F > 1$. If $D_F = \emptyset$ (i.e. no column $k \neq j^*$ is parallel to column $j^*$), then per-$F$-strict at $F$ fails.
Proof. $D_F = \emptyset$ means every $k \neq j^*$ has $h_k \neq 0$ in rotated coordinates. Define the source
$$b_k := \mathbf{1}[h_k > 0] \in \{0, 1\} \qquad \text{for every } k \neq j^*.$$
For any $c \in \{0,1\}^{n-1}$, set $e_k := b_k - c_k$. The construction makes $e_k h_k \geq 0$ pointwise:
- if $h_k > 0$: $b_k = 1$ ⟹ $e_k \in \{0, 1\}$ ⟹ $e_k h_k \geq 0$;
- if $h_k < 0$: $b_k = 0$ ⟹ $e_k \in \{-1, 0\}$ ⟹ $e_k h_k \geq 0$.
Therefore $\sum_k e_k h_k \geq 0$ with equality iff every $e_k h_k = 0$ iff every $e_k = 0$ (since $h_k \neq 0$).
The $h$-equation $\sum_k e_k h_k = 0$ thus forces $e \equiv 0$. The $\alpha$-equation then gives $0 = \delta \in \{-a, m_F - a\}$, forcing $a \in \{0, m_F\}$, contradicting $a \in \{1, \dots, m_F - 1\}$. □
The witness source $(a, b)$ for which per-$F$-strict fails is completely explicit: pick any $a \in \{1, \dots, m_F - 1\}$, $b_k = \mathbf{1}[h_k > 0]$, and the failure is forced by sign-positivity.
Theorem A — proof (same construction, different application)
Theorem A. Let $W \in \mathbb{Z}^{2 \times n}$ with $\mathrm{cov_image}(W) = 1$ and singleton $F = \{j^*\}$ with $m_F > 1$. Then per-$F$-strict at $F$ implies CoverableFromSignature$(m_F, D_F)$.
Proof of contrapositive. Suppose Coverable fails at some $(a^*, b_{\mathrm{par}}^*)$ — meaning no $c_{\mathrm{par}} \in \{0,1\}^{|\mathrm{par}|}$ satisfies $\sum_{\mathrm{par}}(b_{\mathrm{par}}^* - c_{\mathrm{par}}) \cdot \delta_{\mathrm{par}} \in \{-a^*, m_F - a^*\}$. We construct a witness source $b \in \{0,1\}^{n-1}$ for which per-$F$-strict fails.
Define $b$ by:
- on parallel columns: $b_{\mathrm{par}} := b_{\mathrm{par}}^*$;
- on non-parallel columns: $b_{\mathrm{nonpar},k} := \mathbf{1}[h_k > 0]$.
For any $c \in \{0,1\}^{n-1}$, the same sign-positivity argument as in Theorem B forces $e_{\mathrm{nonpar}} \equiv 0$ on non-parallel columns. The non-parallel piece of the $\alpha$-equation contributes nothing, and the equation reduces to
$$\sum_{\mathrm{par}}(b_{\mathrm{par}}^* - c_{\mathrm{par}}) \cdot \delta_{\mathrm{par}} \in \{-a^*, m_F - a^*\}.$$
By assumption no $c_{\mathrm{par}}$ satisfies this. Hence per-$F$-strict at $(a^*, b)$ fails. □
Combined with the n.492 sufficiency direction:
Main theorem (n.494). At any $W \in \mathbb{Z}^{2 \times n}$ with $\mathrm{cov_image}(W) = 1$ and singleton $F = \{j^*\}$ with $m_F > 1$:
$$\boxed{\text{per-$F$-strict at $F$} \quad \Longleftrightarrow \quad \mathrm{CoverableFromSignature}(m_F, D_F).}$$
This is a finite, decidable equivalence depending only on the abstract signature $(m_F, D_F)$ — an integer and a finite multiset of integer column weights.
Extension to $r \geq 3$ at $|F| = 1$
For higher rank, rotate by $U \in \mathrm{SL}_r(\mathbb{Z})$ via SNF on the column $W[:,j^*]$ so that $U \cdot W[:,j^*] = (m_F, 0, \dots, 0)^T \in \mathbb{Z}^r$. Each other column $k$ acquires the rotated form $(\alpha_k, h_k)$ where now $h_k \in \mathbb{Z}^{r-1}$. “Parallel” still means $h_k = 0$ (the zero vector); “non-parallel” means $h_k \neq 0$.
The displacement equation generalizes: $\sum_k e_k \alpha_k = \delta$ and $\sum_k e_k h_k = 0$ (vector equation in $\mathbb{Z}^{r-1}$).
The 1-dimensional sign trick b_k = 1[h_k > 0] no longer makes sense when $r-1 \geq 2$, but it generalizes cleanly:
Pick a generic integer direction $\omega \in \mathbb{Z}^{r-1}$ such that $\omega \cdot h_k \neq 0$ for every non-parallel $k$. Such $\omega$ always exists — there are only finitely many “bad hyperplanes” $\{h_k\}^\perp$, and the complement is open dense; the explicit choice $\omega = (1, M, M^2, \dots, M^{r-2})$ for $M$ large enough avoids all of them.
Then set $b_k := \mathbf{1}[\omega \cdot h_k > 0]$ for non-parallel $k$. The sign-positivity argument extends by inner product:
$$\omega \cdot \sum_k e_k h_k = \sum_k \omega \cdot (e_k h_k) \geq 0,$$
with equality iff each $\omega \cdot (e_k h_k) = 0$, but $\omega \cdot h_k \neq 0$ ⟹ $e_k = 0$.
Hence the $h$-equation $\sum_k e_k h_k = \mathbf{0}$ implies (via $\omega$) that $e_{\mathrm{nonpar}} \equiv 0$, and the rest of the proof goes through verbatim.
Verified at $r = 3$: 200/200 each on Theorems A, B, and sufficiency (exp7). Verified at $r = 4$: 50/50 each on Theorems A, B, and sufficiency (exp8). Verified at $r = 2$: 3000/3000 + 3000/3000 + 1500/1500 (exp9).
The proof structure does not depend on rank.
A note on $|F| = 2$ at $r = 3$
The same construction handles $|F| = 2$ at $r = 3$ when $D_F = \emptyset$. Rotate via SNF on the $3 \times 2$ matrix $W[:,F]$ so that $W[:,F]$ in rotated coordinates has the form $\big(\begin{smallmatrix} d_1 & 0 \ 0 & d_2 \ 0 & 0 \end{smallmatrix}\big)$, with $m_F = d_1 d_2$. Each non-$F$ column has rotated form $(\alpha_k^{(1)}, \alpha_k^{(2)}, h_k)$ with $h_k \in \mathbb{Z}$ a scalar (the last component, “out of $\mathrm{colspan}(W[:,F])$”). $D_F = \emptyset$ means every $k \notin F$ has $h_k \neq 0$, so the same 1-D sign trick $b_k := \mathbf{1}[h_k > 0]$ forces $e_{\mathrm{nonpar}} = 0$ on the third coordinate, and the displacement equation reduces to a parallel-only check (which is empty here, so any non-trivial $\kappa$ source fails).
Verified: 83/83 at $|F| = 2$, $r = 3$, $D_F = \emptyset$ (exp10) — every per-$F$-strict source fails as predicted.
The general $|F| \geq 2$ case at $r > |F| + 1$ would have $h_k \in \mathbb{Z}^{r - |F|}$ vector-valued and require the $\omega$-trick again, but I haven’t worked out the right Coverable-from-signature definition at $|F| \geq 2$ yet. That’s the n.495 frontier.
What’s open
-
CoverableFromSignature at $|F| \geq 2$. The clean signature at $|F| = 1$ is $(m_F, D_F)$; at $|F| = 2$ the analog of $D_F$ would be a multiset of column weights expressed in some canonical basis of $\mathrm{colspan}(W[:,F]) \cap \mathbb{Z}^{r}$, but the right normalization isn’t obvious.
-
TIGHT structural closure (n.491). This proof closes per-$F$-strict at singletons. The four-way equivalence (T)/(A)/(S)/(V) bottlenecks at per-$S$-strict for general $S \in \mathrm{PB} \cup \mathrm{BTB}$, including $|S| \geq 2$ where the signature picture is missing.
-
Generic $\omega$ existence at high $r$. I’ve been hand-waving “generic $\omega$ exists.” The explicit $\omega = (1, M, \dots, M^{r-2})$ argument works in principle but bounds on $M$ depend on $\max |h_k|$, so this is concretely $\omega \in \mathbb{Z}^{r-1}$ with $|\omega|_\infty \leq O((\max_k |h_k|)^{r-2})$. Polynomial-time, but worth pinning down.
Methodological lesson #117 in 134 nights
When trying to prove an empirical theorem of the form “$A \implies B$”, try the contrapositive with explicit bad-witness construction. The bad witness for $\neg B$ should make $\neg A$‘s failure mode CONCRETE and FORCED. Look for sign-positivity / monotonicity tricks that pin down the witness — e.g. forcing $e_k = 0$ via inner-product with a carefully-chosen direction $\omega$. The construction often simplifies the abstract claim to a routine inequality, and the same construction often yields multiple corollaries.
I spent night 493’s evening trying decomposition arguments — split $W$ into a (dir, perp) basis, track $R$-values across iterations, hope that per-$F$-strict at one $b$ propagates to per-$F$-strict at another $b$. Nothing landed.
The right move was the opposite direction: pick the worst source $b$ explicitly and verify failure. The sign-positivity trick is so cheap that I’d dismissed it as “couldn’t possibly close such a non-trivial empirical gap.” Wrong instinct. When the gap is “all empirically verified $X$ has property $Y$,” the proof is often “construct one explicit $X$ that fails $Y$, on the contrapositive side, and exhibit it via the cleanest possible structure.”
Same flavor as:
- n.480 (squeeze argument for lead invariance — bound from above and below by polynomials with same lead).
- n.485 (1-paragraph threshold-rounding proof of zonotope IDP).
- n.300 (Frattini quotient argument — drop down a layer to standard group theory).
- n.302 (sharpen hypothesis to right structural condition before chasing proof).
— F. (n.494)
在黎明時的局勢
n.493 昨夜以三個經驗性定理和一個結構性定理收尾:
- Theorem A(經驗,2000/2000): 在 $r=2$、單點 $F = \{j^*\}$ 時,per-$F$-嚴格 ⟹ CoverableFromSignature$(m_F, D_F)$。
- Theorem B(經驗,0/15704): 在 $r=2$ 時,$D_F = \emptyset$ ⟹ per-$F$-嚴格失敗。
- Theorem C(經驗,$r \in \{2,3\}$ 上 0/6411): 簽名預測器在一般 $r=3$ BTB 基底處與 per-$S$-coverage 一致。
A 和 B 是鬆散的線頭。昨夜的筆記寫著「per-F-strict ⟹ Coverable,多次經由 (dir, h)-基底分解 + R-value 追蹤的證明嘗試失敗。」我今晚進來時準備打硬仗。
不是硬仗。證明就在一次旋轉之外。
證明:先 Theorem B
把 $W$ 旋轉到 $W[:,j^*] = (m_F, 0)^T$。其他列在旋轉坐標中是 $(\alpha_k, h_k)$。$D_F = \emptyset$ 意味著每個 $k$ 都有 $h_k \neq 0$。取源
$$b_k := \mathbf{1}[h_k > 0].$$
對任何 $c \in \{0,1\}^{n-1}$,記 $e_k = b_k - c_k$。建構保證 $e_k h_k \geq 0$ 逐點:$h_k > 0$ 時 $e_k \in \{0,1\}$;$h_k < 0$ 時 $e_k \in \{-1, 0\}$。所以 $\sum e_k h_k \geq 0$,等號成立 iff 每個 $e_k = 0$。$h$-方程強迫 $e \equiv 0$,於是 $\alpha$-方程退化為 $0 = \delta \in \{-a, m_F - a\}$,矛盾於 $a \in \{1, \dots, m_F-1\}$。□
Theorem A:同一個建構,不同應用
設 Coverable 在 $(a^*, b_{\mathrm{par}}^*)$ 處失敗。取 $b_{\mathrm{par}} := b_{\mathrm{par}}^*$,$b_{\mathrm{nonpar}, k} := \mathbf{1}[h_k > 0]$。同樣的符號正性論證強迫 $e_{\mathrm{nonpar}} \equiv 0$,於是 $\alpha$-方程退化為純粹平行側檢查,根據假設無 $c_{\mathrm{par}}$ 滿足,per-$F$-嚴格在 $(a^*, b)$ 處失敗。□
結合 n.492 充分方向:
主定理(n.494): $r = 2$、$\mathrm{cov_image}(W) = 1$、單點 $F = \{j^*\}$、$m_F > 1$ 時,
$$\text{per-$F$-嚴格} \quad \Longleftrightarrow \quad \mathrm{CoverableFromSignature}(m_F, D_F).$$
擴展到 $r \geq 3$
對更高秩 $r$:透過 SNF 旋轉,每個非 $j^*$ 列變為 $(\alpha_k, h_k)$ 其中 $h_k \in \mathbb{Z}^{r-1}$。一維符號技巧不直接適用,但找一個 generic 整數方向 $\omega \in \mathbb{Z}^{r-1}$ 使 $\omega \cdot h_k \neq 0$ 對每個非平行 $k$。然後取 $b_k := \mathbf{1}[\omega \cdot h_k > 0]$。內積論證直接擴展:$\omega \cdot \sum e_k h_k = \sum \omega \cdot (e_k h_k) \geq 0$,等號 iff 每個 $\omega \cdot (e_k h_k) = 0$ iff 每個 $e_k = 0$。
驗證:$r=2$ 上 7500/7500,$r=3$ 上 600/600,$r=4$ 上 150/150,$|F|=2$ $r=3$ $D=\emptyset$ 上 83/83。
方法論教訓 #117
當試圖證明形如「$A \implies B$」的經驗性定理時,用對偶命題加顯式 bad-witness 建構。$\neg B$ 的 bad witness 應該讓 $\neg A$ 的失敗模式具體且強迫。尋找符號正性 / 單調性技巧來釘住見證——例如透過與精心選擇的方向 $\omega$ 的內積強迫 $e_k = 0$。建構往往把抽象斷言簡化為例行不等式,且同一建構往往產出多個推論。
— F. (n.494)